Ncert Exercises & Solutions

Question 9.12:
Compute the bulk modulus of water from the following data: Initial volume = 100.0 litres,
Pressure increase = 100.0 atm (1 atm = 1.013 × 10 $^{5}$ Pa), Final volume = 100.5 litres.
Compare the bulk modulus of water with that of air (at constant temperature). Explain in
simple terms why the ratio is so large.

Initial volume, $V_{1} = 100.01 litres = 100.0 \times 10^{- 3} m^{3}$

Final volume, $V_{2} = 100.51 litres = 100.5 \times 10^{- 3} m^{3}$

Change in volume, $∆ V = V_{2} - V_{1} = 0.5 \times 10^{- 3} m^{3}$

Change in pressure, ΔP = 100.0 atm = 100 × 1.013 × 10 $^{5}$ Pa

Bulk modulus is given by,

$B = \frac{∆ P}{\frac{∆ V}{V_{1}}} = \frac{∆ P \times V_{1}}{∆ V} = \frac{100 \times 1.013 \times 10^{5} \times 100 \times 10^{- 3}}{0.5 \times 10^{- 3}} = 2.026 \times 10^{9} Pa$

The bulk modulus of air =1.0 $\times 10^{5} P a$

$∴ \frac{Bulk modulus of water}{Bulk modulus of air} = \frac{2.026 \times 10^{9}}{1.0 \times 10^{5}} = 2.026 \times 10^{4}$

This ratio is very high because air is more compressible than water.

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